Direct Energy, 2018a

326 14.4 Derivation of the Innitesimal Generators
t → ˜t =
[
t + εt (∂ t t)+ 1 ]
2! ɛ2 t (∂ t t)(∂ t t)+...
(14.67)
The dependent variable transforms as
t → ˜t = te ε (14.68)
y → ỹ = e ε(t∂ t−3y∂ y ) y. (14.69)
[
y → ỹ = 1+ε (t∂ t − 3y∂ y )+ 1 ]
2! ε2 (t∂ t − 3y∂ y ) 2 + ... y (14.70)
y → ỹ = ye −3ε (14.71)
Dening the constant c 6 = e ε , the transformation can be written as
t → c 6 t and y → (c 6 ) −3 y. (14.72)
The analysis above shows that the original Thomas Fermi equation of Eq.
14.46 and the transformed equation
d ( )
2 yc −3
6
d (tc 6 ) 2
= ( )
yc −3 3/2
6 (tc6 ) −1/2 (14.73)
have the same solutions. From it, we can conclude that if y(t) is a solution
to the Thomas Fermi equation, we know that c −3
6 y(τ) for τ = c 6 t is also a
solution.
14.4.3 Line Equation Example
Consider another example of this procedure applied to the equation ÿ =0.
The solution of this equation can be found by inspection
y(t) =c 0 t + c 1 (14.74)
because this is the equation of a straight line. The coecients c n are constants,
and they are dierent from the previous example. In this example,
we will identify the innitesimal generators for continuous symmetries of
this equation, and we will nd eight innitesimal generators. The result of
this problem appears in [191], and it is a modied version of problem 2.26
of reference [164, p. 180].
Solutions of the original equation must be the same as solutions of an
equation transformed by a continuous symmetry, and this idea is contained
in the symmetry condition of Eq. 14.45. In this case, the original equation
is ÿ =0, so the prolongation of an innitesimal generator acting on this